Question

Suppose x has a distribution with μ = 24 and σ = 21.

(a) If a random sample of size n = 34 is drawn, find μ_x, σ_x and P(24 ≤ x ≤ 26). (Round σ_x to two decimal places and the probability to four decimal places.)

μx =

σx =

P(24 ≤ x ≤ 26) =

(b) If a random sample of size n = 71 is drawn, find μ_x, σ_x and P(24 ≤ x ≤ 26). (Round σ_x to two decimal places and the probability to four decimal places.)

μx =

σx =

P(24 ≤ x ≤ 26) =

(c) Why should you expect the probability of part (b) to be higher than that of part (a)? (Hint: Consider the standard deviations in parts (a) and (b).)
The standard deviation of part (b) is  ---Select--- smaller than the same as larger than part (a) because of the  ---Select--- larger same smaller sample size. Therefore, the distribution about μ_x is  ---Select--- the same wider narrower .

Answer 1

a)
Here,

μx = 24,

σx = 21/sqrt(34) = 3.60,

x1 = 24 and x2 = 26. We need to compute P(24<= X <= 26). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (24 - 24)/3.6 = 0
z2 = (26 - 24)/3.6 = 0.56

Therefore, we get
P(24 <= X <= 26) = P((26 - 24)/3.6) <= z <= (26 - 24)/3.6)
= P(0 <= z <= 0.56) = P(z <= 0.56) - P(z <= 0)
= 0.7123 - 0.5
= 0.2123

P(24 ≤ x ≤ 26) = 0.2123

b)
Here,

μx = 24,

σx = 21/sqrt(71) = 2.49,

x1 = 24 and x2 = 26. We need to compute P(24<= X <= 26). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (24 - 24)/2.49 = 0
z2 = (26 - 24)/2.49 = 0.8

Therefore, we get
P(24 <= X <= 26) = P((26 - 24)/2.49) <= z <= (26 - 24)/2.49)
= P(0 <= z <= 0.8) = P(z <= 0.8) - P(z <= 0)
= 0.7881 - 0.5
= 0.2881

P(24 ≤ x ≤ 26) = 0.2881

c)
The standard deviation of part (b) is smaller than part (a) because of the larger sample size. Therefore, the distribution about μx is the narrower.